1(1994), No. 2, 127-140
ON THE UNIQUENESS THEOREMS FOR THE EXTERNAL PROBLEMS OF THE COUPLE-STRESS
THEORY OF ELASTICITY
T. BUCHUKURI AND T. GEGELIA
Abstract. A formula is obtained for the asymptotic representation of solutions of the basic equations of the couple-stress theory of elas-ticity. The formula is used in proving the uniqueness theorems of the external boundary value problems.
0. Let Ω+be a bounded domain in the three-dimensional Euclidean space R3, and Ω− a complement of Ω+to the entire spaceR3:Ω− =R3\Ω−. The
boundary value problems formulated for the domain Ω−are called external. The uniqueness theorems for the external boundary value problems are valid only under some restrictions of the class of solutions at infinity ([1], [2]). These restrictions arose naturally from the Green formulas and consist in the requirement that both the solution and its derivatives vanish at infinity. The weakening of the restrictions is important from both the theoretical and the practical standpoints (for example, in constructing effective solutions). This question is discussed in the monograph [1] devoted specially to uni-queness theorems of the theory of elasticity.
In recent years new results have been obtained for the external static problems of the classical theory of elasticity [3]-[7]. In these works the authors have succeeded in weakening essentially the restrictions at infinity imposed on the class of solutions in which the uniqueness theorems are proved. The results were obtained by two different methods: in [4], [5] the proof was based on Korn’s inequality, whereas in [3], [6], [7] use was made of the asymptotic representation of solutions in the neighborhood of an isolated singular point (in particular, in the neighborhood of the point at infinity). However, both methods were applied to the system of equations containing only derivatives of higher (second) order. The system of static equations of the classical elasticity theory is also such a system.
In this paper we show that the method of asymptotic representations of solutions in the neighborhood of an isolated singular point (see [3], [6], [7])
can be as well applied to systems of equations containing derivatives of both higher and lower orders. This is exemplified by the system of static equa-tions of the couple-stress theory of elasticity for a homogeneous anisotropic medium containing derivatives of second order, as well as derivatives of first and zero orders. Here we have derived the asymptotic representation of the solution of the said system in the neighborhood of the point at infinity, which has enabled us to prove new uniqueness theorems for the external boundary value problems of the couple-stress theory of elasticity.
The derivation of asymptotic representations largely rests on the behavior of the fundamental solution of the considered system at infinity.
1. A homogeneous system of the couple-stress theory of a homogeneous anisotropic micropolar elastic medium is written in the form [2], [9]
cijlk ∂
2uk
∂xj∂xl −cjilmεklm ∂ωk ∂xj = 0, cjmlkεijm∂uk
∂xl +c ′ jilk
∂2ωk
∂xj∂xl−cjmlpεijmεklpωk= 0, i= 1,2,3, (1)
where u = (u1, u2, u3) is the displacement vector, ω = (ω1, ω2, ω3) is the
rotation vector,εijk is the Levi-Civita symbol,cjilk,c′
jilk(i, j, l, k= 1,2,3) are the elastic constants. Here and in what follows the repetition of the index in the product means summation over this index.
It is assumed that the elastic coefficientscijkl andc′
ijkl satisfy the sym-metry conditions
cijkl=cklij, c′ijkl=c′klij (2) and the energetic form is positive-definite
cijklξijξkl+c′ijklηijηkl>0 for ξijξij+ηijηij6= 0. (3) Let
A(∂x)≡ kAik(∂x)k6×6,
Aik(∂x)≡cjilk ∂
2
∂xj∂xl, i, k= 1,2,3; Ai,k+3(∂x)≡ −cjilmεklm ∂
∂xj, i, k= 1,2,3; Ai+3,k(∂x)≡cjmlkεijm ∂
∂xl, i, k= 1,2,3; Ai+3,k+3(∂x)≡c′jilk
∂2
Denote byU = (U1, . . . , U6) the six-component vectorUi =ui and Ui+3=
ωi (i= 1,2,3). Then the system (1) is written in the matrix form
A(∂x)U = 0 Aik(∂x)Uk = 0. (4) 2. Let us establish the properties of the fundamental matrix Φ =kΦijk6×6
of the operatorA(∂x) in the neighborhood of the point at infinity. By virtue of the definition of the fundamental matrix we have
Aik(∂x)Φkj(x) =δijδ(x), i, j= 1, . . . ,6,
where δij is the Kronecker symbol andδ is the Dirac function. Using the Fourier transform
b
ϕ(ξ)≡F[ϕ](ξ) = Z
R3
eix·ξϕ(x)dx,
from this equality we obtain
−Aik(ξ)Φkj(ξ) =b δij, i, j= 1, . . .6, where
A(ξ)≡ kAik(ξ)k6×6,
Aik(ξ) =cjilkξjξl, Ai,k+3(ξ) =icjilmεklmξj,
Ai+3,k(ξ) =−icjmlkεijmξl,
Ai+3,k+3(ξ) =c′jilkξjξl+cjmlpεijmεklp, i, k= 1,2,3.
(5)
The matrix A(ξ) is the invertible one if |ξ| ≡ (ξiξi)1/2 6= 0. Indeed, if |ξ| 6= 0 andηi ≡ξi/|ξ|, then
detA(ξ) =|ξ|6detB(η,|ξ|), B(η, ρ)≡ kBik(η, ρ)k6×6,
Bik(η, ρ) =Aik(η), i≤3 or k≤3;
Bi+3,k+3(η, ρ) =ρ2cjilk′ ηjηl+cjmlpεijmεklp, i≤3, k≤3.
(6)
Now we will prove that detB(η, ρ)6= 0 forη6= 0. Consider the expression Bik(η, ρ)UiUk=cjilkηjηluiuk+ρ2c′
jilkηjηlωiωk+cjmlpεijmεklpωiωk. By virtue of (3) we have the estimates
cjilkηjηluiuk ≥c0(ηjui)(ηjui) =|η|2couiui,
c′
jilkηjηlωiωk≥c0(ηjωi)(ηjωi) =|η|2coωiωi,
for some positive numberc0. Therefore
Bik(η, ρ)UiUk ≥ |η2|c0uiui+|η|2ρ2c0ωiωi+ 2c0ωiωi>0
ifU 6= 0 andη6= 0. Therefore detB(η, ρ)6= 0 forη6= 0. Let us represent detB(η, r) as follows:
detB(η, ρ) =
6
X
k=0
ak(η)ρk,
whereak(η) are homogeneous polynomials ofηof orderk+ 6. In particular, a0(η) = detB(η,0).
As proved above,
a0(η)6= 0, 6
X
k=0
ak(η)ρk6= 0 for η6= 0. (7)
Write detA(ξ) in the form
detA(ξ) =|ξ|6 6
X
k=0
ak(η)|ξ|k. (8)
By virtue of (7)A(ξ) is the invertible matrix for|ξ| 6= 0. Therefore b
Φik(ξ) =−A−ik1(ξ), i, k= 1, . . . ,6.
Let us now estimate the elements of the matrixΦ(ξ). First we will proveb the validity of the representation
b
Φik(ξ) =Φb(1)ik(ξ) +Φb(2)ik(ξ),
i, k= 1, . . . ,6, (9)
where Φb(1)ik(ξ) are homogeneous functions of order −2 for i, k = 1,2,3, of order−1 fori= 1,2,3 andk= 4,5,6, and fori= 4,5,6 andk= 1,2,3; of order 0 fori, k= 4,5,6;Φb(2)ik(ξ) admits the estimates
|∂αΦb(2)ik(ξ)| ≤cα|ξ|−|α|−1, i≤3, k≤3;
|∂αΦb(2)ik(ξ)| ≤cα|ξ|−|α|, i≤3, k≥4 or i≥4, k≤3;
|∂αΦb(2)ik(ξ)| ≤cα|ξ|1−|α|, i≥4, k≥4,
(10)
Let F ≡ (F1, . . . , F6) be some vector and Vi ≡ A−ik1Fk. Repeating the
above arguments for the matrixB(η, r), we can readily prove
Aik(ξ)ViVk ≥c0|ξ|2ViVi+ 2c0 4
X
i=1
Vi2. (11)
Let us fix the indexp. If Fk =δkp,k= 1, . . . ,6, thenVi =A−ik1(ξ)δkp= A−ip1(ξ) (i= 1, . . . ,6). The substitution of the obtained value of Vi in (11) leads to
A−pp1(ξ)≥c0|ξ|2 6
X
k=1
A−kp1(ξ)2+ 2c0 6
X
k=4
A−kp1(ξ)2. (12)
Hence
6
X
k,p=1
A−kp1(ξ)2≤ 1
c0|ξ|2 6
X
p=1
A−pp1(ξ)≤ c1 |ξ|2
X6
k,p=1
A−kp1(ξ)21/2,
|A−kp1(ξ)| ≤c1|ξ|−2, c1= const, k, p= 1, . . . ,6.
(13)
From (12) and (13) we obtain
6
X
k=4
A−kp1(ξ)2≤ 1
2c0
A−1
pp(ξ)≤ c1
2c0 |ξ|−2,
|A−kp1(ξ)| ≤c2|ξ|−1, k= 4,5,6; p= 1, . . . ,6.
(14)
SinceAik(ξ) =Aki(−ξ) (i, k= 1, . . . ,6), from (14) we have
|A−kp1(ξ)| ≤c2|ξ|−1, c1= const, k= 1, . . . ,6; p= 4,5,6. (14′)
Considering (12) forp≥4, we obtain
6
X
k,p=4
A−kp1(ξ)2≤ 1
2c0 6
X
p=4
A−pp1(ξ)≤c1
X6
k,p=4
A−kp1(ξ)21/2.
Therefore
|A−kp1(ξ)| ≤c1, k, p= 4,5,6. (15)
Now we will prove the representation (9). Leti ≤3 and k ≤3. Write b
Φik(ξ) in the form b
Φik(ξ) =−A−ik1(iξ) =− Mik(ξ)
where Mik(ξ) is the cofactor of the element Aik(ξ) in the matrix A(ξ). ThereforeMik(ξ) is the polynomial of ξ. Since detA(ξ) =|ξ|6detB(η,|ξ|)
and|Φik(ξ)b | ≤c|ξ|−2, it is obvious thatMik(ξ) is represented in the form
Mik(ξ) =|ξ|4 6
X
j=0
bikj (η)|ξ|j,
wherebj(η) is the homogeneous polynomial ofηof orderj+4 (j= 1, . . . ,6). Thus
b
Φik(ξ) =− 1 |ξ|2
P6
j=0bikj (η)|ξ|j P6
j=0aj(η)|ξ|j
.
Setting
b
Φ(1)ik(ξ) =− 1 |ξ|2·
bik
0 (η)
a0(η)
=− 1 |ξ|2
bik
0
ξ |ξ|
a0
ξ |ξ|
,
b
Φ(2)ik(ξ) =− 1
|ξ|
P6
j=1(a0(η)bikj (η)−bik0(η)aj(η))|ξ|j−1
a0(η)P6j=0aj(η)|ξ|j
,
(16)
we obtain the required representation (9), since Φb(1)ik(ξ) is a homogeneous function ofξ of order−2 andΦb(2)ik(ξ) satisfies the condition (10) for i, k= 1,2,3.
In a similar manner one can prove the validity of the representation (9) for the rest ofiandk.
Let us now estimate the matrix Φ(x). From the equality (9) we have
Φik(x) = Φ(1)ik(x) + Φ(2)ik(x), i, k= 1, . . . ,6. (17) The first term in (17) is the inverse Fourier transform of the homogeneous functionΦb(1)ik(ξ), and therefore Φ
(1)
ik(ξ) is a homogeneous function of order
−3−q, whereqis the order of the homogeneous functionΦb(1)ik(ξ). Thus for Φ(1)ik(ξ) we have the estimates
|∂αΦ(1)
ik(x)| ≤c|x|−|α|−1, i≤3, k≤3;
|∂αΦ(1)ik(x)| ≤c|x|−|α|−2, i≤3, k≥4 or i≥3, k≤4;
|∂αΦ(1)ik(x)| ≤c|x|−|α|−3, i, k≥4, c= const.
Next we will estimate the second term in (17). It will be shown that in the neighborhood of the point at infinity
∂αΦ(2)ik(x) =o(|x|−|α|−1), i≤3, k≤3;
∂αΦ(2)ik(x) =o(|x|−|α|−2), i≤3, k≥4 or i≥3, k≤4; ∂αΦ(2)
ik(x) =o(|x|−|α|−3), i≥4, k≥4.
(19)
We introduce the functionsω0andω1, whereω1= 1−ω0andω0possesses
the following properties:
ω0∈C∞(R3), suppω0⊂B(0,1), ω0(x) = 1 if |x| ≤
1 2. HereB(0,1) is the ball with center 0 and radius 1 inR3. Obviously,
b
Φ(2)ik(ξ) =Φb(2)ik(ξ)ω0(ξ) +Φb(2)ik(ξ)ω1(ξ),
Φ(2)ik(x) =Φ0(2)ik(x)+
1
Φ(2)ik(x), where
0
Φ(2)ik(x) =F −1[Φb(2)
ikω0](x), 1
Φ(2)ik(x) =F −1[Φb(2)
ikω1](x).
F−1 is the inverse Fourier transform operator.
Leti ≤3,k ≤3 and |β| < α+ 2. Then by virtue of (10) the function ∂β(ξαΦb(2)
ik(ξ)ω0(ξ)) is absolutely integrable onR3and therefore the inverse Fourier transform of this function tends to zero at infinity, but
F−1[∂β(ξαΦb(2)ik(ξ)ω0(ξ))](x) = (−1)|α|i|α|+|β|xβ∂α 0
Φ(2)ik(x). Thus, if|β|=|α|+ 1, then
lim |x|→∞x
β∂α 0
Φ(2)ik(x) = 0 and therefore
∂αΦ0(2)ik(x) =o(|x|
−|α|−1). (20)
Let us estimateΦ1(2)ik(x). Ifn≥ |α|+ 2, then ∆n(ξαΦb(2)ik(ξ)ω1(ξ))∈L1(R3)
Therefore for anyn≥ |α|+ 2
∂αΦ1ik(2)(x) =o(|x|
−2n). (21)
Equations (20) and (21) imply the first estimate in (19). The rest of the estimates are proved in the same manner.
3. The derivation of the asymptotic representation formula for the solution of the system (1) in the neighborhood of the point at infinity is based on the Green formulas. We will give these formulas.
Let Ω be a bounded domain in R3 with a piecewise-smooth boundary ∂Ω,U = (U1, . . . , U6),V = (V1, . . . , V6),U ∈C2( ¯Ω) andV ∈C2( ¯Ω). Then
Z
Ω
(Vi(x)AikUk(x)−Uk(x)Aki(∂x)Vi(x))dx= =
Z
∂Ω
(Vi(y)Tik(∂y, ν)Uk(y)−Uk(y)Tki(∂y, ν)Vi(y))dy S, (22)
where T(∂y, ν)≡ kTik(∂y)k6×6 is the boundary stress operator defined on
∂Ω by the relations
Tik(∂y, ν) =cjilkνj ∂ ∂yl, Ti,k+3(∂y, ν) =−cjilmνjεklm,
Ti+3,k(∂y, ν) = 0,
Ti+3,k+3(∂y, ν) =c′jilkνj ∂
∂yl, i, k= 1,2,3.
(23)
Hereν = (ν1, ν2, ν3) is the unit normal to∂Ω at the pointy, external with
respect to Ω.
If U = (U1, . . . , U6) is the solution of the system (1) in the domain Ω,
belonging to the classC2(Ω)∩C1( ¯Ω), then∀x∈Ω:
uj(x) = Z
∂Ω
(Ui(y)Tik(∂y, ν)Φkj(y−x)−
4. Let us formulate the theorem of the asymptotic representation of a solution of the system (1) in the neighborhood of the point at infinity.
Theorem 1. LetΩbe a domain fromR3 containing the neighborhood of the point at infinity, letU be a solution of the system (1)inΩof the class
C2(Ω), and let
Ui(z) =o(|z|p+1), i= 1, . . . ,6 (25)
in the neighborhood of the point at infinity, wherepis a nonnegative integer number. Then the representation
Uj(x) = X |α|≤p
c(jα)xα+ X |β|≤q
d(kβ)∂βΦjk(x) +ψj(x) j= 1, . . . ,6
(26)
holds in the neigborhood of the point at infinity. Here c(jα) andd(kβ) are the constants, α= (α1, α2, α3) and β = (β1, β2, β3)are the multiindexes, q is an arbitrary nonnegative integer number, and the function ψj admits the estimate
∂γψj(x) =O(|x|−2−|γ|−q), j= 1, . . . ,6 (27)
in the neighborhood of |x|=∞, where γ≡(γ1, γ2, γ3)is an arbitrary mul-tiindex.
Moreover, each of the three terms on the right-hand side of (26)is the solution of the system (1) in the neighborhood of|x|=∞.
Proof. Let x ∈ Ω, and let a positive number r be chosen such that x∈B(0, r/8) andR3\B(0, r/8)⊂Ω. Write the formula (24) for the domain Ωr≡B(0, r)∩Ω. We will have
Uj(x) = Z
∂Ω
Ui(y)Tik(∂y, ν)Φkj(y−x)−
−Φkj(y−x)Tki(∂y, ν)Ui(y)dyS+ +
Z
∂B(0,r)
Ui(y)Tik(∂y, ν)Φkj(y−x)−
Represent Φkj(y−x), in the neighborhood of the pointy, by the Taylor’s formula
Φkj(y−x) = X |α|≤p+1
(−1)|α|xα α! ∂
αΦkj(y) +Rkj(x, y),
Rkj(x, y) = X |α|≤p+2
(−1)|α|xα α! ∂
αΦkj(y−θx), 0< θ <1.
(29)
By virtue of (18) and (19) we readily ascertain that the estimates
|∂yβRkj(x, y)| ≤aβ,p(r)|y|−p−|β|−3, k, j≤3;
|∂yβRkj(x, y)| ≤aβ,p(r)|y|−p−|β|−4, k≤3, j≥4 or k≥4, j≤3;
|∂yβRkj(x, y)| ≤aβ,p(r)|y|−p−|β|−4, k, j≥4
(30)
are fulfilled for any xand y satisfying the conditions|x|< r/8 andr/4≤ |y| ≤r. Taking into account (29), from (28) we obtain
Uj(x) =Uj(0)(x) + X
|α|≤p+1
(−1)|α|c(α) j (r) α! x
α+Ij(p, r, x),
j= 1, . . . ,6,
(31)
where
Uj(0)(x)≡ Z
∂Ω
Ui(y)Tik(∂y, ν)Φkj(y−x)−
−Φkj(y−x)Tki(∂y, ν)Ui(y)dyS, (32) c(jα)(r)≡
Z
∂Ω
Ui(y)Tik(∂y, ν)∂αΦkj(y)−
−∂αΦkj(y)Tki(∂y, ν)Ui(y)dyS, (33) Ij(p, r, x)≡
Z
∂B(0,r)
Ui(y)Tik(∂y, ν)Rkj(x, y)−
−Rkj(x, y)Tki(∂y, ν)Ui(y)dyS. (34) It is not difficult to prove (cf. [6]) thatc(jα)(r) does not depend onr, and, introducing the notation
c(jα)≡
(−1)|α| α! c
(α)
j (r), we obtain the equality
Uj(x) =Uj(0)(x) + X |α|≤p+1
from which we conclude thatIj(p, r, x) does not depend onreither. Thus, if we prove
lim
r→∞Ij(p, r, x) = 0, we will obtain
Uj(x) =Uj(0)(x) + X
|α|≤p+1
c(jα)xα. (35)
Let the function ω:R3→R, ω∈C∞
0 (R3), suppω⊂B(0,3)\B(0,1/3),
ω(y) = 1 for 1/2<|y|<2. Then the estimate
|∂αω(r)(y)| ≤b(α)r−|α| (36) holds for the functionω(r)(y)≡ω(y/r).
Rewriting the formula (22) for the domainB(0, r)\B(0, r/4), in whichV is replaced by the function
V ≡R1(rj)(x,·), . . . , R(6rj)(x,·), R(kjr)(x, y)≡ω(r)(y)Rkj(x, y), we obtain
Ij(p, r, x) = Z
B(0,r)\B(0,r/4)
Ui(z)Aik(∂z)R(kjr)(x, z)dz, j= 1, . . . ,6.
(37)
On account of (30) we have the estimates
|Aik(∂z)Rkj(x, z)| ≤a(x)|z|−p−5, i≤3, j≤3;
|Aik(∂z)Rkj(x, z)| ≤a(x)|z|−p−6, i≤3, j≥4;
|Aik(∂z)Rkj(x, z)| ≤a(x)|z|−p−4, i≥4, j≤3;
|Aik(∂z)Rkj(x, z)| ≤a(x)|z|−p−5, i≥4, j≥4. Taking these estimates and restrictions (25) into account, we obtain
lim
r→∞Ij(p, r, x) = 0.
The representation (35) is thus derived. Note that, due to (25), in the formula (35) the constantsc(jα)= 0 ifα=p+ 1, and therefore we have the representation
Uj(x) =Uj(0)(x) + X |α|≤p
c(jα)xα.
Let us transform this representation in the form (26). To this effect, in the formula (32) we will represent Φkj(y−x) by the Taylor’s formula. Since Akj(ξ) =Ajk(−ξ), we haveA−kj1(ξ) =A
−1
Φjk(x−y). Choose a positive numberr0such thatR3\B(0, r0)⊂Ω. Then,
ify∈∂Ω andx∈R3\B(0,2r0), we will have the expansion
Φkj(y−x) = Φjk(x−y) =
= X
|α|≤q
(−1)|α|yα α! (∂
αΦjk)(x) +ψjk(x, y),
ψkj(x, y) = X |α|=q+1
(−1)q+1yα α! (∂
αΦjk)(x−θy),
0< θ <1.
(38)
Applying the estimates (18), (19), we show that
|∂βxψjk(x, y)| ≤c
(β)
jk (y)|x|
−q−|β|−2,
j≤3, k≤3;
|∂βxψjk(x, y)| ≤c
(β)
jk (y)|x|
−q−|β|−3,
j≤3, k≥4 or j ≥4, k≤3;
|∂βxψjk(x, y)| ≤c
(β)
jk (y)|x|
−q−β−4,
j≥4, k≥4.
(39)
The substitution of (38) in (32) gives Uj(0)(x) =
X
|α|≤q
d(kα)∂αΦjk(x) +ψj(x), (40)
ψj(x) = (−1)q X |α|=q
Z
∂Ω
Ui(y)y α
α!Tik(∂x, ν)(∂
αΦjk)(x)dy S−
−
Z
∂Ω
(Ui(y)Tik(∂x, ν)ψjk(x, y)−ψjk(x, y)Tki(∂y, ν)Ui(y))dy S. Now, due to (18), (19) and (39), we obtain
|∂γψj(x)| ≤c(jγ)|x|−|γ|−2−q, j = 1, . . . ,6.
Remark. Theorem 1 can also be proved when the condition (25) is re-placed by the conditions of Theorem 2 from [6].
In the domain Ω− with the piecewise-smooth boundary∂Ω, find a solu-tionU of the system (1) of the classC1( ¯Ω)∩C2(Ω), satisfying the boundary
condition
∀y ∈∂Ω : lim
Ω−∋x→y
U(x) =ϕ(y)
and the condition at infinity lim
|x|→∞U(x) = 0.
Theorem 2. The first external problem of the couple-stress theory of elasticity has at most one solution.
Proof. LetU be a solution of the first external problem. Then the expansion (26) holds forU. Setting p= 0, q= 0 in (26), we obtain the equality
Uj(x) =c(0)j +d(0)k Φjk(x) +ψj(x), j= 1, . . . ,6.
All terms on the right-hand side of this equality, except c(0)j , tend to zero as |x| → ∞. Thereforec(0)j = 0, j = 1, . . . ,6. Now we conclude from (18), (19), (27) that
∂αUj(x) =O(|x|−|α|−1), j= 1,2,3; ∂αUj(x) =O(|x|−|α|−2), j= 4,5,6.
Now, repeating the arguments, say, from [2], we readily obtain the proof of Theorem 2.
The uniqueness theorems for the other external boundary value problems of the couple-stress elasticity are proved in a similar manner.
References
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Springer tracts in natural philosophy,v. 19,Springer-Verlag, Berlin-Heidel-berg-New York,1971.
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(Received 02.03.1993) Authors’ address:
